A-level Mathematics/Edexcel/Further 1/Proof by Mathematical Induction

=Proof by mathematical induction =

Mathematical induction is the process of verifying or proving a mathematical statement is true for all values of $$n$$ within given parameters. For example:

$$ Prove\ that\ f(n)=5^n+8n+3\ is\ divisible\ by\ 4,for\ n \in \mathbb{Z^+} $$

We are asked to prove that $$ f(n) $$ is divisible by 4. We can test if it's true by giving $$n\ $$values.

So, the first 5 values of n are divisible by 4, but what about all cases? That's where mathematical induction comes in.

Mathematical induction is a rigorous process, as such all proofs must have the same general format:
 * 1) Proposition - What are you trying to prove?
 * 2) basis case - Is it true for the first case? This means is it true for the first possible value of n.
 * 3) Assumption - We assume what we are trying to prove is true for a general number. such as $$ k $$
 * 4) Induction - Show that if our assumption is true for the ($$ k^{th}) $$ term, then it must be true for the term after ($$k+1^{th}) $$term.
 * 5) Conclusion - Formalise your proof.

There will be four types of mathematical induction you will come across in FP1:
 * 1) Summing series
 * 2) Divisibility
 * 3) Recurrence relations
 * 4) Matrices

Example of proofs by Induction involving Summing series
Proposition: $$ Prove\ by\ induction\ that, \displaystyle\sum_{r=1}^n r^3 =\frac{1}{4}n^2(n+1)^2 $$$$, for\ n\in \mathbb{N^+} $$

Note our parameter, $$ for\ n\in \mathbb{N^+} $$ This means it wants us to prove that it's true for all values of $$n$$ which belong to the set($$\in$$) of positive integers ($$\mathbb{N^+} $$)

Basis case:

$$ Let\ n=1\ $$

$$\displaystyle\sum_{r=1}^1 r^3 = 1^3\ [LHS]$$

$$\frac{1}{4}1^2(1+1)^2 = 1\ [RHS] $$

The Left hand side of our equation is equal to the right hand side of our equation, therefore our basis case holds true.

$$ LHS=RHS\longrightarrow \displaystyle\sum_{r=1}^n r^3 =\frac{1}{4}n^2(n+1)^2, for\ n=1 $$

Now you make your assumption:

$$ Now,\ let\ n=k\ and\ assume\ true\ \forall\ k \in \mathbb{Z^+} $$

This is what we are assuming to be true for all values of K which belong to the set of positive integers:

$$\displaystyle\sum_{r=1}^k r^3 =\frac{1}{4}k^2(k+1)^2 $$

Induction: For the induction we need to utilise the fact we are assuming $$ n=k $$ to be true, as such we can just add on another term $$ (k+1)$$ to the summation of the $$ k^{th} $$ term in the series to give us the summation of the $$ (k+1)^{th} $$ term of the series:

$$ Hence,\ let\ n=k+1: \displaystyle\sum_{r=1}^{k+1} r^3 = \displaystyle\sum_{r=1}^k r^3 + (k+1)^3 $$

$$\displaystyle\sum_{r=1}^{k+1} r^3 =\frac{1}{4}k^2(k+1)^2+(k+1)^3 $$

Factoring out $$ \frac{1}{4}(k+1)^2$$ we get: $$\displaystyle\sum_{r=1}^{k+1} r^3 =\frac{1}{4}(k+1)^2[k^2+4(k+1)] $$

This gives us: $$\displaystyle\sum_{r=1}^{k+1} r^3 =\frac{1}{4}(k+1)^2(k+2)^2 $$

Note we know that we're finished because, looking at what we were originally asked to prove, the $$ n $$ values are replaced with $$ k+1 $$.

Summary:

$$\therefore\ true\ for\ n=k+1 \longrightarrow\ true\ \forall\ n \in \mathbb{Z^+} $$

Therefore, if our assumption is true for $$ n=k$$ then $$n=k+1$$ is also true, which implies that $$n$$ is true for all values of $$n$$ that belong to the set of positive integers.

Example of a proof by induction involving Divisibility
Proposition: $$ Prove\ that\ 4\ |\ f(n)=5^n+8n+3,\ for\ n\in \mathbb{N^+} $$

Again, note our parameter, $$ for\ n\in \mathbb{N^+} $$ This means it wants us to prove that it's true for all values of $$n$$ which belong to the set($$\in$$) of positive integers ($$\mathbb{N^+} $$)

Basis case:

$$ Let\ n=1$$

$$f(1) = 5^1+8(1)+3 \Rightarrow f(1) = 16$$

$$\therefore\ 4\ |\ f(1)$$

Assumption: Now we let $$ n=k$$ where $$k$$ is a general positive integer and we assume that $$ 4\ |\ f(k) $$

Remember $$ f(k)=5^k+8k+3$$

Induction: Now we want to prove that the $$k+1^{th}$$ term is also divisible by 4

Hence $$ let\ n = k+1 \rightarrow\ f(k+1) = 5^{k+1}+8(k+1)+3 $$

This is where our assumption comes in, if $$ 4\ |\ f(k) $$then 4 must also divide$$ f(k+1)-f(k) $$

So: $$ f(k+1)-f(k) = 5^{k+1}+8(k+1)+3 -(5^k+8k+3) $$

$$ f(k+1)-f(k) = 5(5^k) -5^k +8 $$ $$ f(k+1)-f(k) = 4(5^k) +8 $$ $$\therefore f(k+1)-f(k) = 4(5^k +2) $$

Now we've shown $$4\ |\ f(k+1) - f(k) $$ and thus $$ 4\ |\ f(k+1) $$ it implies $$ 4\ |\ f(n) $$ because you have successfully shown that 4 divides $$ f(n)$$, where $$ n $$ is a general, positive integer ($$ k$$) and also the consecutive term after the general term ($$k+1 $$)

Conclusion:

If $$ 4\ |\ f(k) \Rightarrow 4\ |\ f(k+1) $$

$$\therefore\ 4\ |\ f(n),\forall\ n \in \mathbb{Z^+} $$

$$If\ 4\ divides\ f(k)\ (as\ we\ assumed)\ then\ it\ is\ implies\ that\ 4\ also\ divides\ f(k+1), $$

$$therefore\ 4\ divides\ f(n),\ for\ all\ values\ of\ n\ that\ belong\ to\ the\ set\ of\ positive\ integers.$$